Question.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the
(i) radius r' of the new sphere,
(ii) ratio of S and S'.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the
(i) radius r' of the new sphere,
(ii) ratio of S and S'.
Solution:
(i)Radius of 1 solid iron sphere $=r$
Volume of 1 solid iron sphere $=\frac{4}{3} \pi r^{3}$
Volume of 27 solid iron spheres $=27 \times \frac{4}{3} \pi r^{3}$
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be $r$ '.
Volume of new solid iron sphere $=\frac{4}{3} \pi r^{3}$
$\frac{4}{3} \pi r^{\prime 3}=27 \times \frac{4}{3} \pi r^{3}$
$r^{\prime 3}=27 r^{3}$
$r^{\prime}=3 r^{\prime}$
(ii) Surface area of 1 solid iron sphere of radius $r=4 \pi r^{2}$
Surface area of iron sphere of radius $r^{\prime}=4 \pi\left(r^{\prime}\right)^{2}$
$=4 \pi(3 r)^{2}=36 \pi r^{2}$
$\frac{\mathrm{S}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{36 \pi \mathrm{r}^{2}}=\frac{1}{9}=1: 9$
(i)Radius of 1 solid iron sphere $=r$
Volume of 1 solid iron sphere $=\frac{4}{3} \pi r^{3}$
Volume of 27 solid iron spheres $=27 \times \frac{4}{3} \pi r^{3}$
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be $r$ '.
Volume of new solid iron sphere $=\frac{4}{3} \pi r^{3}$
$\frac{4}{3} \pi r^{\prime 3}=27 \times \frac{4}{3} \pi r^{3}$
$r^{\prime 3}=27 r^{3}$
$r^{\prime}=3 r^{\prime}$
(ii) Surface area of 1 solid iron sphere of radius $r=4 \pi r^{2}$
Surface area of iron sphere of radius $r^{\prime}=4 \pi\left(r^{\prime}\right)^{2}$
$=4 \pi(3 r)^{2}=36 \pi r^{2}$
$\frac{\mathrm{S}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{36 \pi \mathrm{r}^{2}}=\frac{1}{9}=1: 9$